CBSE Class 11-science Answered
a person standing on ground wants to hit a plane which is moving 72 km per hour horizontal direction at a height of 490 meter calculate the initial velocity and angle of projection on the bullet and gun respectively
Asked by sharmaronakmad | 26 Sep, 2019, 09:52: PM
Expert Answer
In this problem, initial position of the plane with respect to man while shooting the plane is not given explicitly.
Let us assume Initial position of the plane is just above the man , while the man is shooting as shown in figure.
plane travels with velocity 72 km/hr = 72×(5/18) = 20 m/s
if bullet is fired with initial velocity u and the projection angle α with horizontal ,
then the horizontal componenet of velocity ( u cosα ) of bullet should be same as velocity of plane
u cosα = 20 m/s ..................... (1)
if maximum height reached by bullet is same as the vertical distance of the plane above the man,
then we have, ( u2 sin2α )/(2g) = 490 or u sinα = ( 490×2×9.8)1/2 = 98 m/s ......................(2)
Dividing eqn.(2) by eqn.(1) , we get , tanα = 98/20 = 4.9 or α = tan-1 (4.9) = 78.5°
from eqn.(1), we get u = 20 / cos 78.5° ≈ 100 m/s
Answered by Thiyagarajan K | 26 Sep, 2019, 11:11: PM
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